MCQ
The function $f(x)=\frac{x^2+2 x-15}{x^2-4 x+9}, x \in R$ is
- Aboth one-one and onto.
- Bonto but not one-one.
- Cneither one-one nor onto.
- Done-one but not onto.
✓
Answer
Ans. Bonus
$f(x)=\frac{(x+5)(x-3)}{x^2-4 x+9}$
Let $ g(x)=x^2-4 x+9$
$D<0$
$g(x)>0\text { for } x \in R$
$\therefore\left[ f(-5)=0 \ f(3)=0 \right].$
So, $f(x)$ is many $-$ one.
again,
$y x^2-4 x y+9 y=x^2+2 x-15$
$x^2(y-1)-2 x(2 y+1)+(9 y+15)=0$
$\text { for } \forall x \in R $
$\Rightarrow D \geq 0$
$D=4(2 y+1)^2-4(y-1)(9 y+15) \geq 0$
$5 y^2+2 y+16 \leq 0$
$(5 y-8)(y+2) \leq 0$
Note : If function is defined from $f : R \rightarrow R$ then only correct answer is option $(3)$
$\Rightarrow$ Bonus
$f(x)=\frac{(x+5)(x-3)}{x^2-4 x+9}$
Let $ g(x)=x^2-4 x+9$
$D<0$
$g(x)>0\text { for } x \in R$
$\therefore\left[ f(-5)=0 \ f(3)=0 \right].$
So, $f(x)$ is many $-$ one.
again,
$y x^2-4 x y+9 y=x^2+2 x-15$
$x^2(y-1)-2 x(2 y+1)+(9 y+15)=0$
$\text { for } \forall x \in R $
$\Rightarrow D \geq 0$
$D=4(2 y+1)^2-4(y-1)(9 y+15) \geq 0$
$5 y^2+2 y+16 \leq 0$
$(5 y-8)(y+2) \leq 0$
Note : If function is defined from $f : R \rightarrow R$ then only correct answer is option $(3)$
$\Rightarrow$ Bonus
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